Let \(V\) be a complex vector space and \(W\) a subset of \(V\) with the following properties:
\(W\) is a real vector space with respect to the operations defined in \(V\).
For each \(\alpha\) in \(V\) there exist unique vectors \(\beta, \gamma\) in \(W\) such that \(\alpha = \beta + i\gamma\).
Show that the equation \(J\alpha = \beta - i\gamma\) defines a conjugation on \(V\) such that \(J\alpha = \alpha\) if and only if \(\alpha\) belongs to \(W\), and show also that \(J\) is the only conjugation on \(V\) with this property.
Solution
For each \(\alpha\in V\), let \[ \alpha=\beta+i\gamma, \qquad \beta,\gamma\in W \] be the unique decomposition supplied by the hypothesis, and define \[ J\alpha=\beta-i\gamma. \]
1. The map \(J\) is additive and conjugate homogeneous.
If \[ \alpha=\beta+i\gamma, \qquad \alpha'=\beta'+i\gamma', \] with all four vectors in \(W\), then \[ \alpha+\alpha' =(\beta+\beta')+i(\gamma+\gamma'). \] Since \(W\) is a real vector space, uniqueness of the decomposition gives \[ J(\alpha+\alpha') =(\beta+\beta')-i(\gamma+\gamma') =J\alpha+J\alpha'. \]
Now let \(c=a+ib\in\mathbb C\), with \(a,b\in\mathbb R\). Then \[ c\alpha =(a\beta-b\gamma)+i(b\beta+a\gamma), \] and both displayed components lie in \(W\). Therefore \[ \begin{aligned} J(c\alpha) &=(a\beta-b\gamma)-i(b\beta+a\gamma)\\ &=(a-ib)(\beta-i\gamma)\\ &=\overline c\,J\alpha. \end{aligned} \]
2. The map is an involution and its fixed set is exactly \(W\).
If \(\alpha=\beta+i\gamma\), then \[ J\alpha=\beta-i\gamma=\beta+i(-\gamma), \] so applying \(J\) again gives \[ J(J\alpha)=\beta+i\gamma=\alpha. \] Hence \(J^2=I\).
If \(\alpha\in W\), its unique decomposition is \(\alpha=\alpha+i0\), so \(J\alpha=\alpha\). Conversely, if \(J\alpha=\alpha\) and \(\alpha=\beta+i\gamma\), then \[ \beta-i\gamma=\beta+i\gamma, \] so \(2i\gamma=0\), hence \(\gamma=0\) and \(\alpha=\beta\in W\). Thus \[ \boxed{W=\{\alpha:J\alpha=\alpha\}.} \]
3. This conjugation is unique with fixed set \(W\).
Let \(K\) be any conjugation whose fixed set is \(W\). For \(\alpha=\beta+i\gamma\) with \(\beta,\gamma\in W\), \[ K\alpha=K\beta+K(i\gamma) =\beta-i\gamma =J\alpha. \] Thus \(K=J\). Hence the displayed construction is the unique conjugation with fixed set \(W\) [@HK71].